# For any non-empty bounded below set $A$ such that $A \subset \mathbb{R}$, show that $-\inf(A)=\sup(-A)$.

There are various proofs of this on MSE but I didn't think any of them was quite rigorous enough. I've already proved that if $A \subset \mathbb{R}$ is non-empty and bounded below, then it has an infimum.

$A$ is bounded below and non-empty, so $\inf(A)$ exists. Let $\alpha = \inf(A)$. We know that $\alpha \leq x, \forall x \in A$, which is equivalent to knowing that $-\alpha \geq -x, \forall x \in A$. Since $-A$ is the set $\{-x: x \in A\}$, we have proved that $-A$ is bounded above. Then, by the axiom of completeness, $-A$ has a supremum, $\beta = \sup(-A)$ (we have proved existance).

We now want to show that $-\alpha=\beta$. Using the definitions of supremum and infimum, we have that

• $\alpha \leq x, \forall x \in A$,
• If $y \leq x, \forall x \in A$, then $\alpha \geq y$.

and

• $\beta \geq -x, \forall x \in A$,
• If $y \geq -x, \forall x \in A$, then $\beta \leq y$.

We already know that $-\alpha \geq -x, \forall x \in A$, so $\beta \leq -\alpha$. However, we also established that $\beta \geq -x, \forall x \in A$, and $\alpha \in A$. So, $\beta \geq -\alpha$. Then, $\beta = -\alpha$ and the proof is complete.

Are there any logical holes in this proof?